As far as I know, P&L of Plain Vanilla Option depends only on Final price [$] S_T[$]. i.e. P&L is path-independent.
But I see a scenario which P&L is not path-independent, at least for me.
Say, a client want to write Call option on a dividend-paying stock to investment bank.
But this big client don't want to bet on dividend, so he asks the bank to quote option price without forecasted dividend. The client prefer to have his option specification adjusted (treat it like special dividend) once dividend is actually announced.
[$]S_0 = 100[$]
[$]K = 105[$]
[$]T = 1[$]
[$]r = 0\%[$]
[$]\sigma = 20\%[$]
Black-Scholes Pricing gives [$]C1_{open}[$] = 5.90$
So the client receive a premium of 5.90$ from the bank.
Half a year passed, the stock announces to pay a dividend of 5$/share.
However, for some reasons, the client change his mind and don't want to have his option specification adjusted this time.
The client decide to close his position, and simultaneously re-open position (with the same specification as the previous one except that the pricing, this time, incorporate dividend)
The bank agrees but require that the client must complete his plan before the X Date arrives.
At [$]T = 0.5[$]
If the client close position at spot 90
Black-Scholes Pricing gives [$]C1_{close}[$] = 0.96$ (dividend is not incorporated)
the client then re-open position at spot 90
Black-Scholes Pricing gives [$]C2_{open}[$] = 0.56$ (dividend is incorporated)
Therefore, his final P&L is [$]+ C1_{open} - C1_{close} + C2_{open} - max(S_T - K,0)[$]
You may see, from the table, that [$]- C1_{close} + C2_{open}[$] (Net Paid Column) is actually not constant.
So, P&L, somehow, become path-dependent. Level of spot that the client decide to close and re-open position affect its final P&L.
How can I explain this in a layman's term?